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Found in: Page 117

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A 0.150 Kg particle moves along an x axis according to ${\mathbf{x}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{-}}{\mathbf{13}}{\mathbf{.}}{\mathbf{00}}{\mathbf{+}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{t}}{\mathbf{+}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{{\mathbf{t}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{3}}{\mathbf{.}}{\mathbf{00}}{{\mathbf{t}}}^{{\mathbf{3}}}$, with x in meters and t in seconds. In unit-vector notation, what is the net force acting on the particle at t = 3.40 s?

The net force acting on the particle in the unit-vector notation at t = 3.40s is $\left(-7.98\mathrm{N}\right)\stackrel{^}{\mathrm{i}}$ .

See the step by step solution

## Step 1: The given data

• Mass of the particle, m = 0.150 kg.
• Time at which we need to calculate the force, t = 3.40s .
• Position of the particle of mass 0.150 kg along x axis is $\mathrm{x}\left(\mathrm{t}\right)=-13.00+2.00\mathrm{t}+4.00{\mathrm{t}}^{2}-3.00{\mathrm{t}}^{3}$

## Step 2: Understanding the concept of Newton’s law of motion

The net force ${\stackrel{\mathbf{\to }}{\mathbf{F}}}_{{\mathbf{Net}}}$ on a body is a product of mass m and acceleration $\stackrel{\mathbf{\to }}{\mathbf{a}}$of the body and is given by,

${\stackrel{\mathbf{\to }}{\mathbf{F}}}_{\mathbf{N}\mathbf{e}\mathbf{t}}{\mathbf{=}}{\mathbit{m}}\stackrel{\mathbf{\to }}{\mathbf{a}}$

The acceleration can be differentiating the displacement vector twice with respect to time.

Using the formula for Newton’s second law, we can find the net force acting on the particle at t = 3.40 s

Formulae:

The net force acting on the body,

${\stackrel{\to }{F}}_{Net}=M\stackrel{\to }{a}$ (i)

Here, ${\stackrel{\to }{F}}_{net}$is the net force, M is mass of the object, and $\stackrel{\to }{a}$is the acceleration.

The acceleration of a body in motion, $\stackrel{\to }{a}=\left(\frac{{d}^{2}x}{d{t}^{2}}\right)$ (ii)

## Step 3: Calculation of net force

From equation (ii) and the given position equation,

$x\left(t\right)=-13.00+2.00t+4.00{t}^{2}-3.00{t}^{3}$

we get the acceleration of the particle from equation (ii) by substituting the displacement in the equation.

role="math" localid="1657015679628" $\stackrel{\to }{\mathrm{a}}=\left(\frac{{\mathrm{d}}^{2}\left[-13.00+2.00\mathrm{t}+4.00{\mathrm{t}}^{2}-3.00{\mathrm{t}}^{3}\right]}{{\mathrm{dt}}^{2}}\right)\phantom{\rule{0ex}{0ex}}=8.00-18.0\mathrm{t}\phantom{\rule{0ex}{0ex}}=53.2\mathrm{m}/{\mathrm{s}}^{2}\left(\because \mathrm{t}=3.40\mathrm{s}\right)$

Now using equation (i) and the derived and given values, we get the net force as:

${\stackrel{\to }{\mathrm{F}}}_{\mathrm{Net}}=\left(0.150\mathrm{kg}\right)\left(8.00-18.0\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}=\left(-7.98\mathrm{N}\right)\stackrel{^}{\mathrm{i}}\left(\because \mathrm{t}=3.40\mathrm{s}\right)$

Hence, the net force in unit-vector notation is $\left(-7.98\mathrm{N}\right)\stackrel{^}{\mathrm{i}}$.